Gate ECE 2017 Eigen Value

Question

For the given matrix A, one of the Eigenvalue is real

$A=\begin{bmatrix} 1 &2 &3 &4 &5 \\ 5 &1 &2 &3 &4 \\ 4&5 &1 &2 &3 \\ 3&4 &5 &1 &2 \\ 2 &3 &4 &5 &1 \end{bmatrix}$

The real Eigen value is:

Comments

you have already answered it here https://gateoverflow.in/117446/gate-ece-2017-eigen-value

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15 will be the answer.

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For future readers !!!

https://math.stackexchange.com/questions/347408/prove-that-if-the-sum-of-each-row-of-a-equals-s-then-s-is-an-eigenvalue-o

Answer 1

Ans -1

$\begin{bmatrix} 1 & 2 &3 & 4 & 5\\ 5& 1 & 2 & 3 & 4\\ 4& 5 & 1 &2 & 3\\ 3& 4 & 5 & 1 & 2\\ 2& 3 &4 & 5 & 1 \end{bmatrix}$

=$\begin{bmatrix} 15 & 15 &15 & 15 & 15\\ 5& 1 & 2 & 3 & 4\\ 4& 5 & 1 &2 & 3\\ 3& 4 & 5 & 1 & 2\\ 2& 3 &4 & 5 & 1 \end{bmatrix}$

[where $R_{1}\leftarrow R_{1}+R_{2}+R_{3}+R_{4}+R_{5}$]

=$15\begin{bmatrix} 1 & 1 &1 & 1 & 1\\ 5& 1 & 2 & 3 & 4\\ 4& 5 & 1 &2 & 3\\ 3& 4 & 5 & 1 & 2\\ 2& 3 &4 & 5 & 1 \end{bmatrix}$

=$15\begin{bmatrix} 1 & 1 &1 & 1 & 1\\ 14& 13 & 12 & 11 & 10\\ 4& 5 & 1 &2 & 3\\ 3& 4 & 5 & 1 & 2\\ 2& 3 &4 & 5 & 1 \end{bmatrix}$

[Here $R_{2}\leftarrow R_{2}+R_{3}+R_{4}+R_{5}$]

=$15\begin{bmatrix} 0 & 0 & 0 & 0 &1 \\ 1 & 1 &1 &1 & 10\\ -1&4 & -1 &-1 & 3\\ -1 & -1 & 4 &-1 &2 \\ -1 & -1 &-1 &4 & 1 \end{bmatrix}$

[Here $C_{1}\leftarrow C_{1}-C_{2}$

$C_{2}\leftarrow C_{2}-C_{3}$

$C_{3}\leftarrow C_{3}-C_{4}$

$C_{4}\leftarrow C_{4}-C_{5}$]

=$15\times 1\begin{bmatrix} 1 & 1 & 1 &1 \\ -1& 4 & -1 &-1 \\ -1 &-1 &4 &-1 \\ -1& -1 &-1 &4 \end{bmatrix}$

=$15\times 1\begin{bmatrix} 0& 0 & 0 &1 \\ -5& 5 &0 &-1 \\ 0 &-5 &5 &-1 \\ 0& 0 &-5 &4 \end{bmatrix}$

=$15\times 1\times 5\times 5\times 5\begin{bmatrix} -1& 1 &0 \\ 0 &-1 &1 \\ 0& 0 &-1 \end{bmatrix}$

=$15\times 1\times 5\times 5\times 5\begin{bmatrix} -1-\lambda & 1 &0 \\ 0 &-1-\lambda &1 \\ 0& 0 &-1-\lambda \end{bmatrix}$

=(-1-$\lambda$)(-1-$\lambda$)^2

$\lambda =-1$

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How eleminating column

$\begin{bmatrix} 0 &0 & 1\\ x& y &z \\ p& q &r \end{bmatrix}$

=$0\begin{bmatrix} y & z\\ q & r \end{bmatrix}$-$0\begin{bmatrix} x & z\\ p& r \end{bmatrix}$+$1\begin{bmatrix} x & y\\ p& q \end{bmatrix}$

=$1\begin{bmatrix} x & y\\ p& q \end{bmatrix}$

So, 1 row, and 1 column eleminated

Comments

How are you doing this? How columns disappeared?

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the row contain value 1, the matix corresponds to that row . we should calculate it only. rt?

Because $0\begin{bmatrix} x &y \\ w &z \end{bmatrix}$

gives result 0

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I did not get. I cannot go short cut here because I'm not good in linear algebra. But your initial matrix itself looks wrong - how you got it?

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@Srestha I think your mistake is you are applying transformations intially and then substituing the char eqn which wont work.

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ok elaborating

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@Shreshtha : Concept you are applying is 'Eigen values of lower / upper triangular matrix is nothing but their diagonal element' yes it is true.

But  determinant of two matrices are equal doesn't make same eigen values.

Answer 2

Every row sum equals $15$, so $15$ is an eigenvalue.

Comments

Nope

This is not actual procedure

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What do you mean by a actual procedure?

 If you try to expand $det(A-\lambda I )$ then we see  this matrix has four complex eigenvalues (but this is so long)

My answers shows $Ax=15x$ with $x=(1,1,1...,1)^T$

Are  you know this fact? : In a square matrix A with every row sum equals k, then k is an eigenvalue of A